\newproblem{lay:6_1_22}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 6.1.22}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $\mathbf{u}=(u_1,u_2,u_3)$. Explain why $\mathbf{u}\cdot\mathbf{u}\geq 0$. When is $\mathbf{u}\cdot\mathbf{u}=0$?
}{
   % Solution
	\begin{center}
		$\mathbf{u}\cdot \mathbf{u}=u_1^2+u_2^2+u_3^2$
	\end{center}
	Any real number squared is non-negative, and the sum of non-negative numbers is also non-negative. So, $\mathbf{u}\cdot \mathbf{u}\geq 0$.
	$\mathbf{u}\cdot \mathbf{u}=0$ if all its terms are 0, that is, $u_1=u_2=u_3=0$, or what is the same, if $\mathbf{u}=\mathbf{0}$.
}
\useproblem{lay:6_1_22}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
